Please help me understand the voltage regulator with external bypass BJT?

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Such as this schematic:

An externally hosted image should be here but it was not working when we last tested it.


I understand that at low currents the BJT is off and the current is supplied by the voltage regular, then as the current increases a I*R vbe develops on Q1, until the point of Vbe(on) then Q1 starts to conduct.

My question is, how is the feedback maintained to provide constant output voltage? I don't quite understand the feedback path. Vbe should adjust with the load current, but is the current path through the load straight from Vdc to vout to Q1 or is R1 somehow in the path?
 
R2 is a base stopper for the big transistor.
It is there for to make the input to the transistor suitable.
It might work without R2, but we can say R2 is there for safety.

If you look into schematics of power amplifiers, you can see these Base stopper resistors.
They are in small value. 2.2, 4.7 or 10 Ohm.
 
how is the feedback maintained to provide constant output voltage? I don't quite understand the feedback path
It's very simple: look at the current paths.

if output voltage drops due to loading, the regulator tries to maintain the voltage output and that requires it supplies more current into the output. That current must be reflected in the input current to your '7812' input; which increases the voltage across R1 and therefore, Q1 is driven harder.

In effect the expected voltage drop across R1 due to variations in load current is divided by the hfe of the pass transistor.

Q1 remains under control of the 3-pin reg unless cut-off when current through R1 drops low enough that V across r1 <0.6v.
 
I'm sorry to inform you but that schematic and its intended use are incorrect. The circuit in post 1 is a short circuit protection scheme and should not be used as a "current boosted regulator" , for one it will inject all the noise and ripple from the input straight into the output, making the whole point of using a regulator moot, unless you simply want a voltage and dont care if its clean.
Attached is the correct way. Note c1 and c2 are solid tantalum types and should be included with these types of regulators.
 

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Not true at all. Q1 remains under the control of the reg; worst case, the base is driven by the input filtered by R1/C1. The output is quiet!

PS conventional 3-pin regs don't actually need an output cap for stabilty, although it helps output load transient response. LDO regs however usually do require the output cap for stability since it forms part of the reg's compensation.

An input cap is always necessaary though regardless of type.
 
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Load sensitivity of the schematic post 1 will be the one of the regulator
since the transistor is enclosed in a feedback loop while the second schematic
will have poor loading stability , the transistor and current limiting
resistors being outside of the loop.
 
For me, the first pic is the configuration usually associated with the 3-terminal regs. The 2nd pic I usually associate with the uA723 reg. The configurations seem to be somewhat interchangeable.
The feedback network within the 78xx remains functional. If Vout is reduced, that also reduces current through R1.
That's my fingers-crossed take on it.:)
 
the particular circuit is mod as presented in the LM 317 application it is one of the most clever circuits and mother of many others in electronics .

beyond being very quite its extremely tolerant to many loads and that will include a lot of RF applications and you will find similar circuits driving many RF CB amplifiers and many medium power 50-300W FM linear amplifiers ..

Additional features is that the Lm 317 will preserve thermal stability if common heatsinked with the output transistor ( S )

And the trick to preserve 1000% regulation is to remove feedback lead of the LM 317 in this case the output of the LM317 to sense voltage as close to the load possible .

EXAMPLE
suppose you have the schema 1 of the power supply in 13.8 version and a draw of 5 A with 2 transistors ....Suppose also the you used 0.75mm cable to drive the load in a distance of lets say 10m expect to loose almost half of the voltage due to cable loss ...

Simply if the LM 317 senses the voltage on the load will see that voltage is dropped and increase the output until the voltage on the load is exactly as set to 13.8 volts ..

Of course the cable will boil and the intention of this example is understanding how critical feedback might be in circuits like that ...

One may think the combo of the LM317 +2955 as an sziklai amplifier ...there is a lot of juice in this thought and only a few people advanced thoughts like that in the forum ...most of people are focused in the first stages and in EFP outputs ...think people !!!!

Happy regards
Sakis
 
I'm sorry to inform you but that schematic and its intended use are incorrect. The circuit in post 1 is a short circuit protection scheme and should not be used as a "current boosted regulator" , for one it will inject all the noise and ripple from the input straight into the output, making the whole point of using a regulator moot, unless you simply want a voltage and dont care if its clean.
Attached is the correct way. Note c1 and c2 are solid tantalum types and should be included with these types of regulators.

I am sorry, but You are wrong sir.
The cirquit You present here will need to be corrected with an extra diode in the common leg of the regulator IC, as the IC detects the voltage at the output-transistors base. This will give an error to the output voltage that actually varies some with the temperature of the 2N3055 and the current drawn. And the diode in the common leg of the 7812 would only make a minor correction to get the voltage wanted.

The cirquit in the start of this thread will give a much more stable output voltage as the 7812 will sense the voltage directly to the output of the complete regulator.

The ripple-amplification done by the TIP2955 is actually negligeable, and could easily be filtered out.

I agree however of the recommandation of the capasitor 0,33uF on the input of the 7812. I also would remove the 470uF ellyt there.
 
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